Re: 7.4 - TODO : alter table drop foreign key

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От Stephan Szabo
Тема Re: 7.4 - TODO : alter table drop foreign key
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Msg-id 20021205094312.F2765-100000@megazone23.bigpanda.com
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Ответ на Re: 7.4 - TODO : alter table drop foreign key  ("Dan Langille" <dan@langille.org>)
Ответы Re: 7.4 - TODO : alter table drop foreign key  ("Dan Langille" <dan@langille.org>)
Список pgsql-hackers
On Thu, 5 Dec 2002, Dan Langille wrote:

> On 5 Dec 2002 at 9:31, Stephan Szabo wrote:
>
> > When we talk about ALTER TABLE ADD FOREIGN KEY we're being imprecise, so
> > I think that might be why we're talking past each other here.
> >
> > Technically the syntax in question is:
> >  ALTER TABLE <table> ADD <table constraint definition>
> > where CONSTRAINT <name> is an optional leading clause in a table
> > constraint definition.  ADD FOREIGN KEY is a shorthand for a foreign key
> > constraint (technically unnamed).
>
> Understood.
>
> What about allowing a named foreign key?  I haven't checked the RFCs

Here's a part of what SQL92 (draft) has to say about table constraint
definitions:
        <table constraint definition> ::=             [ <constraint name definition> ]             <table constraint> [
<constraintattributes> ]
 
        <table constraint> ::=               <unique constraint definition>             | <referential constraint
definition>            | <check constraint definition>
 

        <constraint name definition> ::= CONSTRAINT <constraint name>
        <referential constraint definition> ::=             FOREIGN KEY <left paren> <referencing columns> <right
paren>              <references specification>
 

11.6     Syntax Rules
        2) If <constraint name definition> is not specified, then a <con-           straint name definition> that
containsan implementation-           dependent <constraint name> is implicit. The assigned <con-           straint
name>shall obey the Syntax Rules of an explicit <con-           straint name>.
 

In our case, the implementation dependent naming scheme is I believe
"$<n>" where <n> is the maximum one already there for that table +1 I
would guess.



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